The wooden beam (E = 10 GPa) is loaded as shown. Using the Method ofSuperposition, find the slope at A and the vertical deflection at C.4 kN4 kNAC◄1.5 m▬▬▬▬1.5 m--3 mB100 m1200 m APPENDIX DEQUATIONS FOR DEFLECTIONS OF SIMPLY SUPPORTED BEAMSSIMPLY SUPPORTED BEAMSLENGTH, Ly₁A.AAdydxdydxMdydxdydxMaxL/2MaxdydxaL/2dydxFy Maxdo9dydx R-У махdxdydxL/2dx/BBBBBXB90dydxdxdydxdxdydxdydxdydxSLOPEMaxBdydxBMaxBA-FL²16EI-Fab(L+ b)6EI LFab(L+ a)6EI LML3EIML6EI,=9,L³24EI,3q,L³128EI,79, L³384EI,dy-79,L³dx 360EI9,L³dx B 45EI,yy =У махy x=aУ махy:y =-Fbxy = (L²-b²-x²)6EI L0≤x≤ay =y ==DEFLECTIONEQUATIONS-Fx48EI,У мах-Mx6EILУмау =-FL³48EL,-Fba6EI,L-(3L² - 4x²)-90x24EI,-qox384EI,=-ML²243EI0≤x≤L/2-(x² -3Lx+2L²)(L²-b²-a²)-59, L¹У мах 384EI,-90L384EI,+17L²x-L³) L/2≤x≤L-90x360EI Lzat x = 0.4226L(x³ - 2Lx² + L³)-(16x³-24Lx² +9L³)0≤x≤L/2-(8x³-24Lx²-6.563×10qL*ELzat x = 0.4598L-(3x4 -10L²x² +7L¹)-6.52 × 10-³q,L¹EI,at x = 0.5193

Given data: P=4 KN =4*103 NThe length L=6 mdistance from A to C ,a=1.5 mdistance from B to…

Answered: The wooden beam (E = 10 GPa) is loaded…

The wooden beam (E = 10 GPa) is loaded as shown. Using the Method of Superposition, find the slope at A and the vertical deflection at C. 4 kN 4 kN A C -1.5 m 1.5 m- + 3 m |В 100 m 200 m

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Concept explainers

Slope and Deflection

The slope in a deflected beam is described as an angle in radians made by the tangent of the section with the original axis of the beam. The deflection at any point on the axis of the beam is defined as the lateral displacement of the point before and after loading.

Question

The wooden beam (E = 10 GPa) is loaded as shown. Using the Method of
Superposition, find the slope at A and the vertical deflection at C.
4 kN
4 kN
A
C
◄1.5 m▬▬▬▬1.5 m-
-3 m
B
100 m
1200 m

APPENDIX D
EQUATIONS FOR DEFLECTIONS OF SIMPLY SUPPORTED BEAMS
SIMPLY SUPPORTED BEAMS
LENGTH, L
y₁
A.
A
A
dy
dx
dy
dx
M
dy
dx
dy
dx
Max
L/2
Max
dy
dx
a
L/2
dy
dx
F
y Max
do
9
dy
dx R
-У мах
dx
dy
dx
L/2
dx/B
B
B
B
B
X
B
90
dy
dx
dx
dy
dx
dx
dy
dx
dy
dx
dy
dx
SLOPE
Max
B
dy
dx
B
Max
B
A
-FL²
16EI
-Fab(L+ b)
6EI L
Fab(L+ a)
6EI L
ML
3EI
ML
6EI,
=
9,L³
24EI,
3q,L³
128EI,
79, L³
384EI,
dy
-79,L³
dx 360EI
9,L³
dx B 45EI,
y
y =
У мах
y x=a
У мах
y:
y =
-Fbx
y = (L²-b²-x²)
6EI L
0≤x≤a
y =
y =
=
DEFLECTION
EQUATIONS
-Fx
48EI,
У мах
-Mx
6EIL
Умау =
-FL³
48EL,
-Fba
6EI,L
-(3L² - 4x²)
-90x
24EI,
-qox
384EI,
=
-ML²
243EI
0≤x≤L/2
-(x² -3Lx+2L²)
(L²-b²-a²)
-59, L¹
У мах 384EI,
-90L
384EI,
+17L²x-L³) L/2≤x≤L
-90x
360EI L
zat x = 0.4226L
(x³ - 2Lx² + L³)
-(16x³-24Lx² +9L³)
0≤x≤L/2
-(8x³-24Lx²
-6.563×10qL*
EL
z
at x = 0.4598L
-(3x4 -10L²x² +7L¹)
-6.52 × 10-³q,L¹
EI,
at x = 0.5193

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